Question
Download Solution PDFA plate at a distance of 0.03 mm from a fixed plate moves at 0.8 m/s and requires a force of 1.50 N/m2 area of plate. Determine the dynamic viscosity of liquid between the plates.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
According to newton's law of viscosity-
This law states that the shear stress (\(\tau\)) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the coefficient of viscosity.
\(\tau=\mu\, .({du\over dy})\)
Where \(\tau=\) shear stress \(={Force\over Area}\)
\(\mu=\) dynamic viscosity
\({du\over dy}=\) velocity gradient
Calculation:
Given data:
Distance between plates (y) = 0.03 mm or 3 × 10-5 m
Velocity of moving plate (\(v\)) = 0.8 m/s
Force per unit area on plate (F/A) = 1.5 N/m2
Dynamic viscosity (μ) =?
\(Shear\, stress(\tau)=\mu× ({v\over y})\)
\(1.5=\mu× ({0.8\over 3\times 10^{-5}})\)
\(\mu={1.5\times 3\times 10^{-5}\over 0.8}\)
\(\mu=5.625\times 10^{-5}\, N-s/m^2\)
\(\mu=56.25\times 10^{-6}\, N.s/m^2\)
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